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Infima of quasi-uniform anti-atoms
Topology and its Applications 153 (2006) 3327–3337 www.elsevier.com/locate/topol
Infima of quasi-uniform anti-atoms ✩ Eliza P. de Jager ∗ , Hans-Peter A. Künzi Department of Mathematics and Applied Mathematics, University of Cape Town, Rondebosch 7701, South Africa Received 26 November 2004; received in revised form 27 July 2005; accepted 27 July 2005
Dedicated to Professor Dr. Keith Hardie on the occasion of his 75th birthday
Abstract Let (q(X), ⊆) denote the lattice consisting of the set q(X) of all quasi-uniformities on a set X, ordered by set-theoretic inclusion ⊆. We observe that a quasi-uniformity on X is the supremum of atoms of (q(X), ⊆) if and only if it is totally bounded and transitive. Each quasi-uniformity on X that is totally bounded or has a linearly ordered base is shown to be the infimum of antiatoms of (q(X), ⊆). Furthermore, each quasi-uniformity U on X such that the topology of the associated supremum uniformity U s is resolvable has the latter property. © 2006 Elsevier B.V. All rights reserved. MSC: 54E15; 54E05; 06B05 Keywords: Quasi-uniformity; Atom; Anti-atom; Totally bounded; Resolvable; Ultrafilter
1. Introduction In the following we continue our study [3] of the set q(X) of all quasi-uniformities on a set X, partially ordered under set-theoretic inclusion ⊆. It is well known that (q(X), ⊆) is a complete lattice [4, p. 2]. We first observe that a quasi-uniformity on X is the supremum of a collection of atoms of (q(X), ⊆) if and only if it is totally bounded and transitive. Then we address the dual and main question dealt with in this article, namely which quasi-uniformities can be represented as the infimum of a collection of anti-atoms of (q(X), ⊆). This question should be compared with the well-known fact that each filter on a set X is the intersection of ultrafilters on X. We will show that many quasi-uniformities on X are the infimum of anti-atoms of (q(X), ⊆). For instance, each nondiscrete quasi-uniformity on X that is generated by a singleton {T } (here T is a preorder on X) is clearly the infimum of the family of all anti-atoms of the form K(x,y) , where K(x,y) = fil{Δ ∪ {(x, y)}} for (x, y) ∈ T \ Δ. In fact, more generally, we shall prove below that each quasi-uniformity on X with a countable (or even linearly ordered) base can be written as the infimum of anti-atoms of (q(X), ⊆). Moreover, each quasi-uniformity U on a set X such that the topology induced by the supremum uniformity U s on X is resolvable is the infimum of anti-atoms of (q(X), ⊆). ✩ This article was completed while the first author was supported by a grant from the National Research Foundation of South Africa. The authors acknowledge support under the bilateral cooperation between Flanders and South Africa (period 2003/4). * Corresponding author. E-mail addresses: [email protected] (E.P. de Jager), [email protected] (H.-P.A. Künzi).
0166-8641/$ – see front matter © 2006 Elsevier B.V. All rights reserved. doi:10.1016/j.topol.2005.07.017
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On the other hand, while it is well known (and easy to see) that there are uniformities on an infinite set X that cannot be written as the infimum of anti-atoms in the lattice of uniformities on X, the authors do not know of a quasi-uniformity on a set X that cannot be written as the infimum of anti-atoms of (q(X), ⊆). Throughout we assume that the reader is familiar with the basic facts regarding quasi-uniformities (see [4,8]), although below we repeat some well-known results to fix our notation and terminology. In recent years the theory of quasi-uniformities has found numerous interesting applications in topological algebra, functional analysis and the theory of hyperspaces and function spaces (see for instance [12,5,11,2]). As usual a transitive and reflexive binary relation on a set X will be called a preorder. By Δ (or ΔX for clarity) we shall denote the diagonal {(x, x): x ∈ X}, and by |X| the cardinality of X. For any subset A of X, SA will denote the preorder [(X \ A) × X] ∪ [A × A]. Given a set X and a subbase S of a filter on X, fil S will denote the filter generated by S on X. By pr1 (respectively pr2 ) we shall denote the projection from the product X × X onto the first (respectively second) factor space X. If U1 and U2 are two quasi-uniformities on a set X and U1 ⊆ U2 , then we shall say that U2 is finer than U1 or that U1 is coarser than U2 . Note that if the intersection of a family of quasi-uniformities on X is a quasi-uniformity, then it is equal to the infimum of that family in (q(X), ⊆). We denote the smallest element of the lattice (q(X), ⊆), namely the indiscrete uniformity, by I. Similarly we shall denote the largest element of the lattice (q(X), ⊆), namely the discrete uniformity, by D. Its induced quasiproximity δD is called the discrete proximity. Quasi-uniformities belonging to the quasi-proximity class of D are called proximally discrete. Quasi-uniformities not belonging to the quasi-proximity class of D are called proximally nondiscrete. Given a quasi-uniformity U on X, then Uω will as usual denote the finest totally bounded quasi-uniformity coarser than U on X. Furthermore, U s = U ∨ U −1 , where U −1 denotes the quasi-uniform conjugate of U . A nonindiscrete quasi-uniformity U on X is called an atom of (q(X), ⊆) if I is the only quasi-uniformity strictly coarser than U . Similarly, a nondiscrete quasi-uniformity U on X is called an anti-atom of (q(X), ⊆) if D is the only quasi-uniformity strictly finer than U . By U|A we shall denote the subspace quasi-uniformity on A ⊆ X of a quasi-uniform space (X, U). A collection of subsets of a quasi-uniform space (X, U) will be called uniformly discrete if there is U ∈ U such that for each x ∈ X, U (x) has a nonempty intersection with at most one subset in the collection. Given a quasi-pseudometric d on a set X (see [4, p. 3]), Ud will denote the induced quasi-pseudometric quasi-uniformity on X. 2. Main results It is known [3, Propositions 1 and 2] that a quasi-uniformity is an atom of (q(X), ⊆) if and only if it is equal to fil{SA } on X × X for some nonempty proper subset A of X. In the light of this result, it is easy to characterize those quasi-uniformities on X that can be written as the supremum of a family of atoms. Proposition 1. A quasi-uniformity on a set X is the supremum of atoms of (q(X), ⊆) if and only if it is totally bounded and transitive. (Here, as usual, ∅ = I.) Proof. Since clearly both transitivity and total boundedness of quasi-uniformities are preserved under the supremum operation (compare [4]), and since each atom of (q(X), ⊆) is totally bounded and transitive [3, Propositions 1 and 2], the supremum of any family of atoms is a totally bounded and transitive quasi-uniformity. On the other hand, if U is a totally bounded and transitive quasi-uniformity on X, then {SA : AδU (X \ A); ∅ ⊂ A ⊂ X} yields a subbase for U [4, Theorem 1.33], which obviously is written as the union of atoms. (Here δU denotes the negation of the quasi-proximity relation induced by U on the power set of X.) Hence U is the supremum of atoms of (q(X), ⊆). 2 Below the authors present some results dealing with the following much more delicate dual question. Problem 1. Which quasi-uniformities on a set X can be written as the infimum of a family of anti-atoms of (q(X), ⊆)?
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Definition 1. Let U be a quasi-uniformity on a set X. We set UK equal to the infimum of all anti-atoms of (q(X), ⊆) finer than U . (By convention, ∅ = D and hence DK = D. Note that by Zorn’s Lemma UK is a nondiscrete quasiuniformity on X whenever U is nondiscrete.) Remark 1. (a) Obviously (UK )K = UK . Furthermore, for elements U and V of q(X), U ⊆ V implies that UK ⊆ VK . (b) Moreover, (U −1 )K = (UK )−1 . In particular, if U is a uniformity, then UK is also a uniformity. (So each antiatom U of the lattice of uniformities on X satisfies U = UK . More generally it follows that each uniformity U on X that is the infimum of anti-atoms of the lattice of uniformities on X satisfies U = UK .) (c) For anyfamily of quasi-uniformities (Ui )i∈I on X such that (Ui )K = Ui whenever i ∈ I , we have that ( i∈I Ui )K = i∈I Ui . Proof. (a) These statements immediately follow from the definition of UK . (b) It is easy to see that if K is an anti-atom of (q(X), ⊆) finer than U , then K−1 is an anti-atom finer than U −1 . The result follows, since (q(X), ⊆) commute, as was noted the conjugation and infimum operations in in [3, Introduction]. (c) We have ( i∈I Ui )K ⊆ (Uj )K whenever j ∈ I . Then ( i∈I Ui )K ⊆ j ∈I (Uj )K = j ∈I Uj ⊆ ( j ∈I Uj )K . Therefore i∈I Ui = ( j ∈I Uj )K . 2 A quasi-uniformity U on X will be called nonsymmetric if it is not a uniformity on X, that is, if U = U −1 . Corollary 1. Suppose that U is a nonsymmetric quasi-uniformity such that U = UK . Then U ∧ U −1 = (U ∧ U −1 )K . Proof. This is a consequence of Remarks 1(b) and 1(c).
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In the following we give some examples of classes of quasi-uniformities U on X for which U = UK . Our method of proof will be based on an idea explained in the next proposition. A general characterization of those quasi-uniformities U on a set X such that U = UK will be given in the next section (see Lemma 3). Let U1 and U2 be two quasi-uniformities on a set X. Then U1 and U2 will be called ∨-complementary provided that U1 ∨ U2 = D. Proposition 2. Let U be a quasi-uniformity on X. Then each quasi-uniformity V on X that is ∨-complementary to UK is ∨-complementary to U . (In particular, each complement of UK is a complement of U in the lattice (q(X), ⊆).) Proof. Suppose otherwise. Then there is a quasi-uniformity V on X such that V ∨ UK = D, but V ∨ U = D. By Zorn’s Lemma there is an anti-atom K of (q(X), ⊆) such that V ∨ U ⊆ K. Thus UK ⊆ K. But then D = V ∨ UK ⊆ K—a contradiction. We conclude that the first statement holds. The second statement immediately follows, since U ⊆ UK . 2 Definition 2. Let T be a preorder on X and let U be a quasi-uniformity on X. We say that T is U -slim provided that there is U ∈ U such that U ∩ T = Δ. Corollary 2. Let U be a quasi-uniformity on X. Then each preorder T on X that is UK -slim is U -slim. Proof. Otherwise the quasi-uniformity V = fil{T } would be ∨-complementary to UK , but not to U , contradicting Proposition 2. 2 Recall that a subset A of a quasi-uniform space (X, U) is called discrete if there is an entourage U ∈ U such that U ∩ (A × A) = ΔA (equivalently, if the subspace quasi-uniformity U|A is discrete). Corollary 3. Let U be a quasi-uniformity on X. Then UK and U have the same discrete subsets in X. Proof. Suppose that there are A ⊆ X and V ∈ UK such that V ∩ (A × A) = ΔA , but that U ∩ (A × A) = ΔA whenever U ∈ U . Set T = Δ ∪ (A × A). Then T is a preorder on X that is UK -slim but not U -slim. We have reached a contradiction to Corollary 2 and conclude that the statement holds. 2
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Corollary 4. Let U be a quasi-uniformity on X. Then for any disjoint A, B ⊆ X and any V ∈ UK there is U ∈ U such that U ∩ (A × B) ⊆ V . Proof. Suppose not. Set T = Δ ∪ [(A × B) \ V ]. Since A and B are disjoint, T is a preorder on X. It is UK -slim but not U -slim by our assumption, contradicting Corollary 2. 2 Corollary 5. Let U be a quasi-uniformity on X. Then UK belongs to the quasi-proximity class of U . (In particular, U and UK induce the same topology.) Proof. Otherwise there are A, B ⊆ X and V ∈ UK such that V ∩ (A × B) = ∅, but U ∩ (A × B) = ∅ whenever U ∈ U . In particular, A and B are disjoint. We have reached a contradiction to Corollary 4 and conclude that UK belongs to the quasi-proximity class of U . 2 Proposition 3. Each totally bounded quasi-uniformity U on a set X is the infimum of anti-atoms of (q(X), ⊆). Proof. In order to reach a contradiction we suppose that U ⊂ UK . We conclude by Corollary 5 that UK cannot be totally bounded, since only the coarsest quasi-uniformity of a quasi-proximity class is totally bounded [4, Theo−1 rem 1.33]. Hence either UK or UK is not hereditarily precompact [7, Lemma 1.1]. Consequently there are V ∈ UK and a sequence (xn )n∈ω in X such that xk ∈ / V (xn ) whenever n, k ∈ ω and n < k (or n > k). Define T1 = Δ ∪ {(xn , xk ): n k; n, k ∈ ω} (or T2 = Δ ∪ {(xn , xk ): n k; n, k ∈ ω}). Then T1 (or T2 ) is a UK -slim preorder. Note that T1 and T2 are not U -slim: Consider U ∈ U . There is a finite cover {Af : f ∈ F } of X such that Af × Af ⊆ U whenever f ∈ F . Hence there are e ∈ F and i, j ∈ ω such that i < j and xi , xj ∈ Ae . It follows that (xi , xj ) ∈ U ∩ T1 (and (xj , xi ) ∈ U ∩ T2 ). We have reached a contradiction to Corollary 2 and conclude that U = UK . 2 Indeed let us note that the preceding argument essentially shows the following: Corollary 6. If U is a quasi-uniformity on X that is hereditarily precompact, then UK is hereditarily precompact, too. Proof. Otherwise the proof above shows how to construct a preorder T on X that is UK -slim but not U -slim, which contradicts Corollary 2. 2 A quasi-uniformity U on X is called proximally fine if it is the finest quasi-uniformity inducing the quasiproximity δU . (But recall that in general a given quasi-proximity class need not possess a finest member [4, p. 25].) Corollary 7. Each proximally fine quasi-uniformity U on X is the infimum of anti-atoms of (q(X), ⊆). In particular, the fine quasi-uniformity of any topological space X is the infimum of anti-atoms of (q(X), ⊆). Proof. If U ⊂ UK and U is proximally fine, then U and UK must induce distinct quasi-proximities, which contradicts Corollary 5. 2 Corollary 8. Let U be a uniformity on X. Then the Hausdorff uniformities determined by UK and U induce the same topology on the set P0 (X) of nonempty subsets of X. Proof. Note that UK is a uniformity, because U is a uniformity. Since U ⊆ UK and since U and UK have the same discrete subsets and belong to the same quasi-proximity class by Corollaries 3 and 5, our assertion follows from a statement of Ward [13, Theorem 1]. 2 Lemma 1. Let U be a quasi-uniformity on X and suppose that V ∈ UK \ U . Let A ⊆ X be of minimal cardinality such that (A × A) ∩ (U \ V ) = ∅ whenever U ∈ U . If B ⊆ A and [(A × B) ∪ (B × A)] ∩ (U \ V ) = ∅ whenever U ∈ U , then we have |B| = |A|. (Note that A cannot be finite.)
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Proof. Assume first that A is finite. Then there must exist a point (a, b) ∈ A × A such that (a, b) ∈ ( U) \ V , since U is a filter. Consequently T = Δ ∪ {(a, b)} is a preorder on X that is UK -slim but not U -slim—a contradiction to Corollary 2. So A is necessarily infinite. In order to reach a contradiction, suppose now that there is a B ⊆ A such that |B| < |A| and [(A × B) ∪ (B × A)] ∩ (U \ V ) = ∅ whenever U ∈ U . By our assumption on A, there is a U0 ∈ U such that (B × B) ∩ (U0 \ V ) = ∅. Since (A × B) ∪ (B × A) = (B × B) ∪ (B × (A \ B)) ∪ ((A \ B) × B), we have that either (B × (A \ B)) ∩ (U \ V ) = ∅ or ((A \ B) × B) ∩ (U \ V ) = ∅ whenever U ∈ U . Let us consider the first case. (The second case can be treated similarly.) Set T = Δ ∪ [(B × (A \ B)) \ V ]. Then T is a preorder on X that is UK -slim but not U -slim—a contradiction to Corollary 2. We deduce that the statement holds. 2 Remark 2. Let us note that in Lemma 1 the infinite cardinal |A| is necessarily at most equal to the minimal cardinality of a base of the subspace quasi-uniformity U|A on A. Proposition 4. Each quasi-uniformity U on X with a linearly ordered base is the infimum of anti-atoms of (q(X), ⊆). In particular, each quasi-uniformity on X with a countable base is the infimum of anti-atoms of (q(X), ⊆). Proof. Let {Uβ : β < δ} be a decreasing base of U (that is, the sequence (Uβ )β<δ is well-ordered by inverse settheoretic inclusion). We can assume that the cardinal δ is equal to one, countably infinite or an uncountable regular cardinal (compare [10]). Suppose that there is V ∈ UK \ U . Let Y ⊆ X be of minimal cardinality such that (Y × Y ) ∩ (Uβ \ V ) = ∅ whenever β ∈ δ. As noted in the preceding lemma, Y is necessarily infinite. It also follows from Remark 2 that certainly |Y | δ. Indeed under our linearity condition, |Y | δ too: Suppose otherwise, that is, suppose |Y | < δ. Then δ is uncountable. Because Y is infinite, for each (y1 , y2 ) ∈ (Y × Y ) \ V there is a minimal β < δ, say β(y1 ,y2 ) , such that / Uβ(y1 ,y2 ) . Since sup{βz : z ∈ (Y × Y ) \ V } < δ by regularity of δ, we have reached the contradiction that (y1 , y2 ) ∈ V ∩ (Y × Y ) ∈ U|Y . Hence |Y | = δ. We now define two sequences (xβ )β∈δ and (yβ )β∈δ of points in Y inductively as follows: Assume that for some γ < δ and all β < γ we have already chosen xβ and yβ so that Aγ = {xβ : β < γ } and Bγ = {yβ : β < γ } are disjoint. Note that this condition of disjointness holds at limit ordinals provided that it holds at all the preceding ordinals. / Bγ , yγ ∈ / Aγ and Suppose that we can find (xγ , yγ ) ∈ [(Uγ \ V ) ∩ (Y × Y )] \ [(Bγ × Y ) ∪ (Y × Aγ )]. Then xγ ∈ xγ = yγ . Set Aγ +1 = {xβ : β < γ + 1} and Bγ +1 = {yβ : β < γ + 1}. Note that Aγ +1 and Bγ +1 are disjoint. The induction would stop if for some γ ∈ δ we had (Uγ \ V ) ∩ (Y × Y ) ⊆ [(Bγ × Y ) ∪ (Y × Aγ )]. But this clearly implies that (U \ V ) ∩ [[(Aγ ∪ Bγ ) × Y ] ∪ [Y × (Aγ ∪ Bγ )]] = ∅ whenever U ∈ U . Since Aγ ∪ Bγ has cardinality smaller than δ, the latter property cannot hold by definition of Y and Lemma 1. We conclude that this case cannot occur so that we can always continue the induction for all γ < δ. Finally we set T = Δ ∪ {(xγ , yγ ): γ < δ}. By the disjointness of Aδ and Bδ , T is a preorder on X. It is UK -slim but not U -slim—a contradiction to Corollary 2. Hence we have shown that U = UK . 2 A topological space X is called resolvable [6] if it has two disjoint dense subsets. Proposition 5. Each quasi-uniformity U on X such that τ (U s ) is resolvable is the infimum of anti-atoms of (q(X), ⊆). Proof. Let D1 , D2 be disjoint dense sets of (X, τ (U s )). In order to reach a contradiction suppose that there is a V ∈ UK such that U \ V 3 = ∅ whenever U ∈ U . Set T = Δ ∪ [(D1 × D2 ) \ V ]. Note that T is a preorder on X, since D1 and D2 are disjoint. Let us show that T is not U -slim: We shall prove that U 3 ∩ T = Δ whenever U ∈ U . Fix U ∈ U . There is (x, y) ∈ U \ V 3 . Then [V (x) × V −1 (y)] ∩ V = ∅. Since according to Corollary 5 the quasi-uniformities UK and U belong to the same quasi-proximity class and thus both induce the bitopological space (X, τ (U), τ (U −1 )), we can choose (d1 , d2 ) ∈ [[U −1 (x) ∩ V (x)] × [V −1 (y) ∩ U (y)]] ∩ [D1 × D2 ]. Then we have (d1 , x) ∈ U , (y, d2 ) ∈ U and (x, y) ∈ U . Therefore (d1 , d2 ) ∈ U 3 \ V . Hence (d1 , d2 ) ∈ U 3 ∩ T . We conclude that T is not U -slim. On the other hand, note that T is UK -slim, since clearly V ∩ T = Δ. We have reached a contradiction to Corollary 2 and conclude that U = UK . 2
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Proposition 6. Let U be a quasi-uniformity on X and let d be a quasi-pseudometric on X such that Ud ⊆ U , but Ud−1 U . Then there is an anti-atom K of (q(X), ⊆) such that U ⊆ K, but Ud−1 K and thus Ud−1 UK . Proof. For each n ∈ ω ∪ {−1} we set Qn = {(x, y) ∈ X × X: d(x, y) 2−n }. Without loss of generality we can assume that Q−1 / U. −1 ∈ For each n ∈ ω set Fn = Δ ∪ (Qn \ Q−1 −1 ). Define n = Fn0 ◦ · · · ◦ Fnk : k ∈ ω; n0 , . . . , nk ∈ ω; 2−n0 + · · · + 2−nk 2−n . F 2 n and F n whenever n ∈ ω. Consider V = fil{F n : n ∈ ω} ∨ U . Note that the quasi Observe that Fn ⊆ F n+1 ⊆ F n and uniformity V on X is distinct from the discrete quasi-uniformity because for each Q ∈ U , Q ∩ Fn ⊆ Q ∩ F −1 −1 Q ∩ Fn = Δ, since Q−1 ∈ / U and thus (Q ∩ Qn ) \ Q−1 = ∅. Therefore by Zorn’s Lemma there exists an anti-atom K of (q(X), ⊆) such that V ⊆ K. In particular, U ⊆ K. In order to see that Ud−1 K, we shall show that Q−1 / K. To this end, it will be sufficient to verify that F 0 ∩ Q−1 0 ∈ 0 = Δ, since Δ ∈ / K and F 0 ∈ K. In order to reach a contradiction assume that F 0 ∩ Q−1 0 = Δ: Then there are s ∈ ω \ {0} and xi ∈ X (i = 0, . . . , s) such that
(1) x0 = xs ;
s−1 −n j 1; and (2) for each j = 0, . . . , s − 1, (xj , xj +1 ) ∈ Qnj \ Q−1 j =0 2 −1 where nj ∈ ω and (3) (x0 , xs ) ∈ Q−1 0 .
Without loss of generality we could assume that xj = xj +1 . −1 By the triangle inequality it follows that (x0 , xs−1 ) ∈ Q0 . Hence (xs−1 , x0 ) ∈ Q−1 0 . Together with (x0 , xs ) ∈ Q0 −1 we conclude that (xs−1 , xs ) ∈ Q−1 −1 . Since also (xs−1 , xs ) ∈ Qns−1 \ Q−1 we have reached a contradiction. Hence −1 indeed F 0 ∩ Q0 = Δ. We have shown that there is an anti-atom K of (q(X), ⊆) such that U ⊆ K and Ud−1 ⊆ K. Therefore Ud−1 ⊆ UK . 2 Corollary 9. Let U be a quasi-uniformity on X. Then U −1 ∧ U = U −1 ∧ UK . Proof. Obviously U −1 ∧ U ⊆ U −1 ∧ UK . In order to prove the converse, recall that U −1 ∧ UK , as any quasiuniformity (see [4, Lemma 1.5]), can be written as the supremum of a family, say (Ud )d∈D , of quasi-pseudometric quasi-uniformities. Let d ∈ D. Then Ud ⊆ U −1 ∧ UK . If Ud ⊆ U , then applying Proposition 6 to U −1 we see that Ud−1 (U −1 )K and thus Ud UK —a contradiction. Hence Ud ⊆ U and Ud ⊆ U −1 ∧ U . Thus U −1 ∧ UK ⊆ U −1 ∧ U and the assertion is verified. 2 Corollary 10. Let U be a uniformity on X. Then any quasi-uniformity V on X such that U ⊆ V ⊆ UK is a uniformity. Proof. Suppose that there is a quasi-uniformity V on X such that V = V −1 and U ⊆ V ⊆ UK . Then U ⊆ V −1 ⊆ UK , because UK is a uniformity. But by Proposition 6 applied to V there is an anti-atom K on X such that V ⊆ K, but V −1 K. Then U ⊆ K and therefore UK ⊆ K. Consequently V −1 ⊆ UK ⊆ K—a contradiction. We conclude that the statement holds. 2 Corollary 11. Let U be any nonsymmetric quasi-uniformity on X and let V be a quasi-uniformity on X such that U ⊆ V ⊆ UK . Then V is nonsymmetric. Proof. Suppose otherwise, that is, that V = V −1 . Let K be any anti-atom of (q(X), ⊆) such that U ⊆ K. Then UK ⊆ K and hence V ⊆ K. Consequently V ⊆ K−1 and thus U ⊆ K−1 and U −1 ⊆ K. Applying Proposition 6 to U , we note that we have reached a contradiction and conclude that V is nonsymmetric. 2
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3. Some further results Remark 3. In [9, p. 7] it was pointed out that any separated uniformity U on an infinite set X with at least one nonisolated point x0 is not the infimum of anti-atoms in the lattice of uniformities on X: Clearly T = Δ ∪ [(X \ {x0 }) × (X \ {x0 })] is contained in any (uniform) anti-atom K finer than U although it certainly does not belong to U . (Indeed suppose that K is an anti-atom finer than U in the lattice of uniformities on X such that T ∈ / K. Then by maximality of K, there is a symmetric entourage K ∈ K such that K 2 ∩ [(X \ {x0 }) × (X \ {x0 })] = Δ. Thus K(x0 ) \ {x0 } can / U . But then contain at most one point, say a. Since U is separated, there is a symmetric U ∈ U such that (x0 , a) ∈ Δ = K ∩ U ∈ K—a contradiction.) On the other hand, Proposition 4 for instance shows that U = UK in the case that such a U has a linearly ordered base. Remark 4. Remark 3 above shows that an analogue to Corollary 5 does not hold in the lattice of uniformities on X. A potential way to construct an example of a quasi-uniformity U on X such that U ⊂ UK is sketched in our next observation. Example 1. Let U be a maximal element among the quasi-uniformities on X that do not contain some fixed preorder T . Then U is either an anti-atom of (q(X), ⊆) or is strictly coarser than UK . Proof. Note that U = D. If U is not an anti-atom of (q(X), ⊆), then any anti-atom K of (q(X), ⊆) finer than U contains T . Therefore T belongs to UK . Hence U and UK are distinct. 2 Problem 2. Is U as defined in Example 1 always an anti-atom of (q(X), ⊆)? Our next results will show a connection between our problem and concepts of finite dimensionality. Remark 5. Let U be a quasi-uniformity on X. Suppose that V ∈ UK \ U . Then (X × X) \ V cannot be written as the union of finitely many transitive relations. In particular, V cannot be written as the intersection of finitely many linear orders. Proof. We suppose that (X × X) \ V = i∈n Si where n ∈ ω and each Si is transitive. Set Ti = Δ ∪ Si for i ∈ n. Since U \ V = ∅ whenever U ∈ U , there is j ∈ n such that Tj ∩ (U \ V ) = ∅ whenever U ∈ U . Thus Tj is not U -slim, although it is UK -slim—a contradiction to Corollary 2. For the second statement, suppose that V = i∈n Li where each Li is a linear order on X. Then V = (X × X) \
i∈n Li where each Li = (X × X) \ Li is transitive, which is impossible by the result just proved. 2 In the preceding remark, V cannot be of the form {(x, y) ∈ X ×X: |f (x)−f (y)| < ε} for some function f : X → R and ε > 0, since clearly V = (X × X) \ (Z+ ∪ Z− ) where Z+ = {(x, y) ∈ X × X: f (x) f (y) + ε} and Z− = {(x, y) ∈ X × X: f (x) f (y) − ε}, and these two latter relations are transitive. Lemma 2. Let C = {Ca : a ∈ I } be a collection of pairwise disjoint subsets of a set X (with linearly ordered index set (I, )). Set A = C. Consider T0 = Δ ∪ {Ca × Ca : a ∈ I }. Then T0 is the complement of the union of six transitive relations on X. Proof. We assume that X is linearly ordered by . Set S0 = (X \ A) × A, S1 = A × (X \ A), S2 = {(x, y) ∈ (X \ A) × S = {Ca × (X \ A): x ≺ y}, S3 = {(x, y) ∈ (X \ A) × (X \ A): y ≺ x}, S4 = {Ca × Ca : a < a ; a, a ∈ I } and 5 Ca : a < a ; a, a ∈ I }. Then each Si where i ∈ 6 is a transitive relation on X and T0 = (X × X) \ ( i∈6 Si ). 2 Proposition 7. Let U be a quasi-uniformity on X, and let d be a quasi-pseudometric on X such that Ud ⊆ UK , but / U . (Here we set Un = {(x, y) ∈ X × X: d(x, y) < 2−n } and Uns = Un ∩ Un−1 whenever n ∈ ω.) U0 ∈ Suppose that each uniform cover of the uniform space (X, Uds ) has a refinement that consists of finitely many (uniformly) Uds -discrete collections. Then Ud−1 UK .
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Proof. Assume that X is linearly ordered. Consider the uniform cover {U2s (x): x ∈ X} of the uniform space (X, Uds ). By our assumption it has a refinement that can be written as the union of finitely many Uds -discrete collections Ci , where i ∈ n. We assume that these collections are linearly ordered. For each i ∈ n there is a ji ∈ ω \ 3 such that Ujsi ∈ Uds and Ujsi witnesses the uniform discreteness of Ci . Hence for each i ∈ n, {Ujsi (C): C ∈ Ci } is a collection of pairwise disjoint subsets of X. Set Ei = Δ ∪ {Ujsi (C) × Ujsi (C): C ∈ Ci } whenever i ∈ n. Then for each i ∈ n, the relation Ei is the complement in X × X of a union of six transitive relations on X by Lemma 2. Set U = i∈n Ujsi . We have U ⊆ i∈n Ei , since i∈n Ci is a cover of X. Furthermore, i∈n Ei ⊆ U0s ⊆ U0 . Hence i∈n Ei does not belong to U . Since the intersection of transitive relations is transitive, we conclude by the aforementioned structure of the complement of each Ei that i∈n Ei is the complement of a finite union of transitive relations on X. Thus it cannot belong to UK \ U according to Remark 5. It follows that i∈n Ei and thus U does not belong to UK , which implies the statement, because Ud ⊆ UK . 2 Corollary 12. Let U be a uniformity on a set X and let T be a preorder on X such that T ∈ / U . Then T ∈ / UK . Proof. Suppose T ∈ UK . Recall that since U is a uniformity, UK is a uniformity. Hence by our assumption fil{T ∩ T −1 } ⊆ UK . For the quasi-pseudometric d on X defined by d(x, y) = 0 if (x, y) ∈ T , and d(x, y) = 1 otherwise, we have U0 = T and Ud = fil{T }. However, then Ud−1 ⊆ UK which clearly contradicts Proposition 7, since the cover {(T ∩ T −1 )(x): x ∈ X} is a partition of X. Thus T ∈ / UK . 2 We next formulate a general characterization of those quasi-uniformities U on X that satisfy the equation U = UK . Lemma 3. A quasi-uniformity U on a set X is the infimum of anti-atoms in the lattice of quasi-uniformities on X if and only if for each quasi-pseudometric d on X such that Ud ⊆ U there exists a quasi-pseudometric d on X such that Ud ∨ U = D and Ud ∨ Ud = D. Proof. Suppose that the given condition holds. In order to reach a contradiction, assume that U ⊂ UK . Then there is a quasi-pseudometric d on X such that Ud ⊆ U , but Ud ⊆ UK . Therefore, by the stated condition there is a quasipseudometric d on X such that Ud ∨ U = D and Ud ∨ Ud = D. We can find an anti-atom K of (q(X), ⊆) such that Ud ∨ U ⊆ K. Then we have UK ⊆ K. Since D = Ud ∨ Ud ⊆ K, we have reached a contradiction. We conclude that the condition implies that U = UK . In order to establish the converse, assume that the stated condition does not hold. Then there is a quasi-pseudometric d on X such that Ud ⊆ U but that for any quasi-pseudometric d on X such that Ud ∨ U = D, we have that Ud ∨ Ud = D. Consider an arbitrary anti-atom K of (q(X), ⊆) such that U ⊆ K. In order to reach a contradiction, we assume that K ∨ Ud = D. This means that there is a quasi-pseudometric d on X such that Ud ⊆ K, Ud ∨ Ud = D and Ud ∨ U = D—a contradiction. Therefore K ∨ Ud = D and thus Ud ⊆ K, since K is an anti-atom. We have shown that Ud ⊆ UK . Thus U ⊂ UK . We conclude that U = UK implies the given condition. Hence the statement of the lemma holds. 2 Remark 6. Note that the lemma above also yields a characterization of those uniformities on a set X that are infima of families of anti-atoms in the lattice of uniformities on X: Just replace quasi-pseudometrics everywhere by pseudometrics in its formulation. Remark 7. It seems natural to distinguish between the proximally discrete and the proximally nondiscrete anti-atoms when studying Problem 1 (see [3]). For instance, it is a trivial observation that for any quasi-uniformity U on X the infimum of all proximally discrete anti-atoms of (q(X), ⊆) above it is finer than Dω . Remark 8. In the following, for any filters F and G on X, we define the quasi-uniformity UF ,G = fil{Δ∪(F ×G): F ∈ F, G ∈ G} on X. It follows from [3, Section 3] that for each anti-atom K of (q(X), ⊆) there are uniquely defined ultrafilters F and G on X such that F ∩ G = ∅ and UF ,G ⊆ K. (Note that in [3] these ultrafilters F and G were denoted by pr1 K and pr2 K .)
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Fix now some ultrafilters F and G on X such that F ∩ G = ∅. If U is a quasi-uniformity on X, then there are anti-atoms finer than UF ,G above it if and only if U ∨ UF ,G = D. In particular, each quasi-uniformity U = D that is finer than UF ,G clearly has only anti-atoms K of (q(X), ⊆) above it that satisfy pr1 K = F and pr2 K = G. Example 2. Let x ∈ X and F = fil{{x}}, and let G be a free ultrafilter on X. Then UF ,G ∧ UG ,F ⊂ UG ,G . Is known [3, Theorem 1] that UF ,G and UG ,F are anti-atoms of (q(X), ⊆), but that UG ,G is not an anti-atom of (q(X), ⊆) [3, Corollary 5]. Note that if U is an anti-atom of the lattice of uniformities on X that contains the uniformity UG ,G , then any anti-atom K of (q(X), ⊆) containing U satisfies UG ,G ⊆ K ∧ K−1 , where the latter infimum is equal to U . Proposition 8. Each quasi-uniformity U on X that is finer than UF ,G , where F and G are distinct ultrafilters on X, is the infimum of anti-atoms of (q(X), ⊆). Proof. There exists A ⊆ X such that A ∈ F and X \ A ∈ G. Let V ∈ UK . By Corollary 4 there is U ∈ U such that U ∩ [Δ ∪ (A × (X \ A))] ⊆ V . But Δ ∪ (A × (X \ A)) ∈ U . Thus V ∈ U and we have shown that U = UK . 2 Problem 3. Does Proposition 8 also hold if F and G are the same free ultrafilter on X? (It follows from Proposition 10 below and Remark 1(c) that an affirmative answer to this question would yield a positive solution to our Problem 1.) Proposition 9. Let U be a quasi-uniformity on a set X. Then U ∨ Dω = set of ultrafilters on X.
F (U
∨ UF ,F ), where F runs through the
Proof. The inclusion ⊆ is obvious. Let V ∈ F (U ∨ UF ,F ). Then for each ultrafilter F on X there are UF ∈ U and AF ∈ F such that UF ∩ (Δ ∪ (AF × AF )) ⊆ V . If {X \ AF : F is an ultrafilter on X} has the finite intersection property, this collection is contained in some ultrafilter G on X; however, this means that ∅ = (X \ AG ) ∩ AG ∈ G and we have reached a contradiction. Thus there is a finite collection E of ultrafilters F on X such that F ∈E AF = X. Consequently ( F ∈E UF ) ∩ F ∈E (AF × AF ) ⊆ V . Therefore V ∈ U ∨ Dω . We conclude that the stated equality holds. 2 Proposition 10. Let U be a quasi-uniformity on a set X. Then U = the set of ultrafilters on X.
(F ,G ) (U
∨ UF ,G ), where F and G run through
Proof. The inclusion ⊆ is evident. Let V ∈ (F ,G ) (U ∨ UF ,G ). According to Proposition 9 there are U0 ∈ U and a finite cover {Ci : i ∈ n} (where n ∈ ω) of X such that U0 ∩ ( i∈n (Ci × Ci )) ⊆ V . Clearly, without loss of generality we can assume that {Ci : i ∈ n} is a partition of X. Furthermore, for each pair (F, G) of distinct ultrafilters F and G on X there are U(F ,G ) ∈ U and disjoint sets F(F ,G ) ∈ F and G(F ,G ) ∈ G such that [Δ ∪ (F(F ,G ) × G(F ,G ) )] ∩ U(F ,G ) ⊆ V . Fix F . Determine i ∈ n such that Ci ∈ F and denote it by CF . Clearly HF := fil{X \ G(F ,G ) : G is an ultrafilter distinct from F on X} ⊆ F , because F(F ,G ) ⊆ X \ G(F ,G ) whenever G is an ultrafilter distinct from F on X. If HF ⊂ F , then HF ⊆ G0 for some ultrafilter G0 on X distinct from F , which leads to the contradiction that Therefore ( G ∈EF X \ G(F ,G ) ) ⊆ CF for some finite subset EF of (X \ G(F ,G0 ) ) ∩ G(F ,G0 ) ∈ G0 . Thus HF = F . ultrafilters G distinct from F on X. Set LF = ( G ∈EF F(F ,G ) ) ∩ CF . Consider now {X \ LF : F is an ultrafilter on X}. If the latter collection has the finite intersection property, then it is contained in an ultrafilter F on X. But then (X \ LF ) ∩ LF ∈ F —a contradiction. Thus there finite collection I is a of ultrafilters F on X such that F ∈I LF = X. Note now that X × X = [ i∈n (Ci × Ci )] ∪ [ F ∈I ( G ∈EF (F(F ,G ) × G(F ,G ) ))]: Indeed let (x, y) ∈ X × X. Then x ∈ Ci and y ∈ Cj for some i, j ∈ n. If i = j , then (x, y) ∈ Ci × Ci . So let us consider the case where i = j . Of course, x ∈ LF for some F ∈ I and so x ∈ Ci ∩ CF and thus Ci = CF . Furthermore y∈ / Ci = CF . Consequently y ∈ G(F ,G ) for some G ∈ EF . We conclude that (x, y) ∈ F(F ,G ) × G(F ,G ) where F ∈ I and G ∈ EF . Hence our auxiliary set-theoretic equality is proved. From this equality and our definition of the entourages U0 and U(F ,G ) it now follows that U0 ∩ F ∈I ( G ∈EF U(F ,G ) ) ⊆ V . Therefore V ∈ U . We conclude that the statement holds. 2
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Corollary 13. Let U be a quasi-uniformity on X. Then U = F (U ∨ UF ,F ) ∧ {K: K is a proximally nondiscrete anti-atom of (q(X), ⊆) finer than U}, where F runs through the set of ultrafilters on X. Proof. This assertion follows from the fact that any anti-atom finer than UF ,G with distinct ultrafilters F and G on X is proximally nondiscrete (see [3, Section 3]) and Propositions 8 and 10 proved above. 2 In the light of the difficulty of Problem 1 it seems appropriate to look for a not too large set QX of quasi-uniformities on X that contains each anti-atom of (q(X), ⊆) as a member and where we can prove that each quasi-uniformity on X can be written as the infimum of a subset of QX . Our next result shows that such a natural set QX indeed exists. In order to define the set QX we need the following definitions: Let H be any ultrafilter on X × X. Define the filter as a subset. Of course, = {Δ ∪ H : H ∈ H} on X × X. Let (H) q be the finest quasi-uniformity on X contained in H H q = D if and only if Δ ∈ H. (H) / H. Finally set QX equal to the set of Now set HX equal to the set of all the ultrafilters H on X × X such that Δ ∈ q where H belongs to HX . all quasi-uniformities of the form (H) Remark 9. Using the notation just introduced, we remark that it is known [3, Theorem 1] that if H is an ultrafilter on is a proximally nondiscrete anti-atom of (q(X), ⊆). If that condition regarding X × X such that pr1 H = pr2 H, then H will necessarily be a quasi-uniformity. the projections is not satisfied, there seems to be no reason to believe that H q. But note that for any ultrafilter H on X × X such that Δ ∈ / H, we have Upr1 H,pr2 H ⊆ (H) Proposition 11. (a) Each quasi-uniformity U on a set X can be represented as the intersection of quasi-uniformities belonging to QX . q , where H is any ultrafilter finer than the filter K = (b) Each anti-atom K of (q(X), ⊆) is of the form (H) fil{K \ Δ: K ∈ K} on X × X. Hence K ∈ QX . Proof. (a) Since Consider the filter U = fil{U \ Δ: U ∈ U} on X × X. ∧∅ = D, we can assume that U is nondiscrete.
H ∈ HX such that U ⊆ H} an ultrafilter on X × X such that U ⊆ H}. Therefore U = {H: Thus U = {H: H is
and consequently U = {(H)q : H ∈ HX such that U ⊆ H}. q for any ultrafilter H on X × X such that K ⊆ H by (b) If K is an anti-atom of (q(X), ⊆), then clearly K = (H) the maximality of K. 2 q )K hold for each ultrafilter H on X × X such that Δ ∈ q = ((H) / H? Problem 4. Does (H) (Because of Proposition 11(a) and Remark 1(c) an affirmative answer to this question would yield a positive solution to Problem 1.) Finally we note that the studied property of quasi-uniformities is preserved under finite unions, and we then apply this result to our Proposition 5 dealing with resolvability that we have presented in the preceding section. Proposition 12. Let U be a quasi-uniformity on a set X. Suppose that A ⊆ X such that (U|A)K = U|A and (U|(X \ A))K = U|(X \ A). Then U = UK . Proof. For any B, C ⊆ X, we set UB,C = U ∨ fil{Δ ∪ (B × C)}. Note first that U = UA,A ∩ UA,X\A ∩ UX\A,A ∩ UX\A,X\A . This equality is established in a similar fashion as the equalities in the proofs of Propositions 9 and 10 with the help of the fact that X × X = (A × A) ∪ (A × (X \ A)) ∪ ((X \ A) × A) ∪ ((X \ A) × (X \ A)). Applying Corollary 4 to the quasi-uniformity UA,X\A and the disjoint subsets A and X \ A of X we immediately see that (UA,X\A )K = UA,X\A . Similarly we conclude that (UX\A,A )K = UX\A,A . Note next that UA,A |A = U|A. Consider any quasi-pseudometric d on X such that Ud UA,A . Then there is m ∈ ω such that [U ∩ (A × A)] \ Um = ∅ whenever U ∈ U . (Here, as before, Un = {(x, y) ∈ X × X: d(x, y) < 2−n } whenever n ∈ ω.) Therefore Ud |A ⊆ UA,A |A. Since (U|A)K = U|A, by Lemma 3 there is a quasi-pseudometric q on A that is bounded by 1 such that Uq ∨ (Ud |A) = D|A and Uq ∨ (U|A) = D|A. Set s(x, y) = q(x, y) if x, y ∈ A, s(x, y) = 0 if x = y and x ∈ X \ A, and s(x, y) = 1 otherwise. Then s is a quasi-pseudometric on X. Furthermore, Us ∨ Ud = D and
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Us ∨ UA,A = D. We conclude that (UA,A )K = UA,A by Lemma 3. Similarly one shows that (UX\A,X\A )K = UX\A,X\A . The statement then follows from Remark 1(c). 2 A topological space that is not resolvable is called irresolvable. It is known that each topological space is the union of two disjoint subspaces C and U where C is closed and resolvable and U is open and hereditarily irresolvable, that is, every nonempty subspace of X is irresolvable [1, p. 292]. Corollary 14. A quasi-uniformity U on X satisfies U = UK if it satisfies U|U = (U|U )K , where U is the hereditarily irresolvable part in the aforementioned representation of the topological space (X, τ (U s )). Proof. By Proposition 5 we have (U|C)K = U|C where C is the resolvable part of the space (X, τ (U s )) mentioned in the representation above. The result follows from Proposition 12 and our hypothesis regarding U|U . 2 Problem 5. For i ∈ {0, 1}, let Ui be two quasi-uniformities on a set X such that Ui = (Ui )K . Does U0 ∨ U1 = (U0 ∨ U1 )K hold? References [1] G. Bezhanishvili, R. Mines, P.J. Morandi, Scattered, Hausdorff-reducible, and hereditarily irresolvable spaces, Topology Appl. 132 (2003) 291–306. [2] J. Cao, I.L. Reilly, S. Romaguera, Some properties of quasi-uniform multifunction spaces, J. Austral. Math. Soc. (Ser. A) 64 (1998) 169–177. [3] E.P. de Jager, H.-P.A. Künzi, Atoms, anti-atoms and complements in the lattice of quasi-uniformities, Topology Appl. 153 (16) (2006) 3140– 3156. [4] P. Fletcher, W.F. Lindgren, Quasi-Uniform Spaces, Dekker, Basel, 1982. [5] L.M. García-Raffi, S. Romaguera, E.A. Sánchez-Pérez, The bicompletion of an asymmetric normed linear space, Acta Math. Hungar. 97 (2002) 183–191. [6] E. Hewitt, A problem of set-theoretic topology, Duke Math. J. 10 (1943) 309–333. [7] H.-P.A. Künzi, Functorial admissible quasi-uniformities on topological spaces, Topology Appl. 43 (1992) 27–36. [8] H.-P.A. Künzi, Nonsymmetric distances and their associated topologies: About the origins of basic ideas in the area of asymmetric topology, in: C.E. Aull, R. Lowen (Eds.), Handbook of the History of General Topology, vol. 3, Kluwer, Dordrecht, 2001, pp. 853–968. [9] J. Pelant, J. Reiterman, V. Rödl, P. Simon, Ultrafilters on ω and atoms in the lattice of uniformities, I and II, Topology Appl. 30 (1988) 1–17 and 107–125. [10] H.-C. Reichel, Basis properties of topologies compatible with (not necessarily symmetric) distance-functions, Gen. Topology Appl. 8 (1978) 283–289. [11] J. Rodríguez-López, S. Romaguera, The relationship between the Vietoris topology and the Hausdorff quasi-uniformity, Topology Appl. 124 (2002) 451–464. [12] S. Romaguera, M. Sanchis, M. Tkachenko, Free paratopological groups, in: Proceedings 17th Summer Conference on Topology and its Applications, Topology Proc. 27 (2003) 613–640. [13] A.J. Ward, On H -equivalence of uniformities (the Isbell–Smith problem), Pacific J. Math. 22 (1967) 189–196.
Infima of quasi-uniform anti-atoms ✩ Eliza P. de Jager ∗ , Hans-Peter A. Künzi Department of Mathematics and Applied Mathematics, University of Cape Town, Rondebosch 7701, South Africa Received 26 November 2004; received in revised form 27 July 2005; accepted 27 July 2005
Dedicated to Professor Dr. Keith Hardie on the occasion of his 75th birthday
Abstract Let (q(X), ⊆) denote the lattice consisting of the set q(X) of all quasi-uniformities on a set X, ordered by set-theoretic inclusion ⊆. We observe that a quasi-uniformity on X is the supremum of atoms of (q(X), ⊆) if and only if it is totally bounded and transitive. Each quasi-uniformity on X that is totally bounded or has a linearly ordered base is shown to be the infimum of antiatoms of (q(X), ⊆). Furthermore, each quasi-uniformity U on X such that the topology of the associated supremum uniformity U s is resolvable has the latter property. © 2006 Elsevier B.V. All rights reserved. MSC: 54E15; 54E05; 06B05 Keywords: Quasi-uniformity; Atom; Anti-atom; Totally bounded; Resolvable; Ultrafilter
1. Introduction In the following we continue our study [3] of the set q(X) of all quasi-uniformities on a set X, partially ordered under set-theoretic inclusion ⊆. It is well known that (q(X), ⊆) is a complete lattice [4, p. 2]. We first observe that a quasi-uniformity on X is the supremum of a collection of atoms of (q(X), ⊆) if and only if it is totally bounded and transitive. Then we address the dual and main question dealt with in this article, namely which quasi-uniformities can be represented as the infimum of a collection of anti-atoms of (q(X), ⊆). This question should be compared with the well-known fact that each filter on a set X is the intersection of ultrafilters on X. We will show that many quasi-uniformities on X are the infimum of anti-atoms of (q(X), ⊆). For instance, each nondiscrete quasi-uniformity on X that is generated by a singleton {T } (here T is a preorder on X) is clearly the infimum of the family of all anti-atoms of the form K(x,y) , where K(x,y) = fil{Δ ∪ {(x, y)}} for (x, y) ∈ T \ Δ. In fact, more generally, we shall prove below that each quasi-uniformity on X with a countable (or even linearly ordered) base can be written as the infimum of anti-atoms of (q(X), ⊆). Moreover, each quasi-uniformity U on a set X such that the topology induced by the supremum uniformity U s on X is resolvable is the infimum of anti-atoms of (q(X), ⊆). ✩ This article was completed while the first author was supported by a grant from the National Research Foundation of South Africa. The authors acknowledge support under the bilateral cooperation between Flanders and South Africa (period 2003/4). * Corresponding author. E-mail addresses: [email protected] (E.P. de Jager), [email protected] (H.-P.A. Künzi).
0166-8641/$ – see front matter © 2006 Elsevier B.V. All rights reserved. doi:10.1016/j.topol.2005.07.017
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On the other hand, while it is well known (and easy to see) that there are uniformities on an infinite set X that cannot be written as the infimum of anti-atoms in the lattice of uniformities on X, the authors do not know of a quasi-uniformity on a set X that cannot be written as the infimum of anti-atoms of (q(X), ⊆). Throughout we assume that the reader is familiar with the basic facts regarding quasi-uniformities (see [4,8]), although below we repeat some well-known results to fix our notation and terminology. In recent years the theory of quasi-uniformities has found numerous interesting applications in topological algebra, functional analysis and the theory of hyperspaces and function spaces (see for instance [12,5,11,2]). As usual a transitive and reflexive binary relation on a set X will be called a preorder. By Δ (or ΔX for clarity) we shall denote the diagonal {(x, x): x ∈ X}, and by |X| the cardinality of X. For any subset A of X, SA will denote the preorder [(X \ A) × X] ∪ [A × A]. Given a set X and a subbase S of a filter on X, fil S will denote the filter generated by S on X. By pr1 (respectively pr2 ) we shall denote the projection from the product X × X onto the first (respectively second) factor space X. If U1 and U2 are two quasi-uniformities on a set X and U1 ⊆ U2 , then we shall say that U2 is finer than U1 or that U1 is coarser than U2 . Note that if the intersection of a family of quasi-uniformities on X is a quasi-uniformity, then it is equal to the infimum of that family in (q(X), ⊆). We denote the smallest element of the lattice (q(X), ⊆), namely the indiscrete uniformity, by I. Similarly we shall denote the largest element of the lattice (q(X), ⊆), namely the discrete uniformity, by D. Its induced quasiproximity δD is called the discrete proximity. Quasi-uniformities belonging to the quasi-proximity class of D are called proximally discrete. Quasi-uniformities not belonging to the quasi-proximity class of D are called proximally nondiscrete. Given a quasi-uniformity U on X, then Uω will as usual denote the finest totally bounded quasi-uniformity coarser than U on X. Furthermore, U s = U ∨ U −1 , where U −1 denotes the quasi-uniform conjugate of U . A nonindiscrete quasi-uniformity U on X is called an atom of (q(X), ⊆) if I is the only quasi-uniformity strictly coarser than U . Similarly, a nondiscrete quasi-uniformity U on X is called an anti-atom of (q(X), ⊆) if D is the only quasi-uniformity strictly finer than U . By U|A we shall denote the subspace quasi-uniformity on A ⊆ X of a quasi-uniform space (X, U). A collection of subsets of a quasi-uniform space (X, U) will be called uniformly discrete if there is U ∈ U such that for each x ∈ X, U (x) has a nonempty intersection with at most one subset in the collection. Given a quasi-pseudometric d on a set X (see [4, p. 3]), Ud will denote the induced quasi-pseudometric quasi-uniformity on X. 2. Main results It is known [3, Propositions 1 and 2] that a quasi-uniformity is an atom of (q(X), ⊆) if and only if it is equal to fil{SA } on X × X for some nonempty proper subset A of X. In the light of this result, it is easy to characterize those quasi-uniformities on X that can be written as the supremum of a family of atoms. Proposition 1. A quasi-uniformity on a set X is the supremum of atoms of (q(X), ⊆) if and only if it is totally bounded and transitive. (Here, as usual, ∅ = I.) Proof. Since clearly both transitivity and total boundedness of quasi-uniformities are preserved under the supremum operation (compare [4]), and since each atom of (q(X), ⊆) is totally bounded and transitive [3, Propositions 1 and 2], the supremum of any family of atoms is a totally bounded and transitive quasi-uniformity. On the other hand, if U is a totally bounded and transitive quasi-uniformity on X, then {SA : AδU (X \ A); ∅ ⊂ A ⊂ X} yields a subbase for U [4, Theorem 1.33], which obviously is written as the union of atoms. (Here δU denotes the negation of the quasi-proximity relation induced by U on the power set of X.) Hence U is the supremum of atoms of (q(X), ⊆). 2 Below the authors present some results dealing with the following much more delicate dual question. Problem 1. Which quasi-uniformities on a set X can be written as the infimum of a family of anti-atoms of (q(X), ⊆)?
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Definition 1. Let U be a quasi-uniformity on a set X. We set UK equal to the infimum of all anti-atoms of (q(X), ⊆) finer than U . (By convention, ∅ = D and hence DK = D. Note that by Zorn’s Lemma UK is a nondiscrete quasiuniformity on X whenever U is nondiscrete.) Remark 1. (a) Obviously (UK )K = UK . Furthermore, for elements U and V of q(X), U ⊆ V implies that UK ⊆ VK . (b) Moreover, (U −1 )K = (UK )−1 . In particular, if U is a uniformity, then UK is also a uniformity. (So each antiatom U of the lattice of uniformities on X satisfies U = UK . More generally it follows that each uniformity U on X that is the infimum of anti-atoms of the lattice of uniformities on X satisfies U = UK .) (c) For anyfamily of quasi-uniformities (Ui )i∈I on X such that (Ui )K = Ui whenever i ∈ I , we have that ( i∈I Ui )K = i∈I Ui . Proof. (a) These statements immediately follow from the definition of UK . (b) It is easy to see that if K is an anti-atom of (q(X), ⊆) finer than U , then K−1 is an anti-atom finer than U −1 . The result follows, since (q(X), ⊆) commute, as was noted the conjugation and infimum operations in in [3, Introduction]. (c) We have ( i∈I Ui )K ⊆ (Uj )K whenever j ∈ I . Then ( i∈I Ui )K ⊆ j ∈I (Uj )K = j ∈I Uj ⊆ ( j ∈I Uj )K . Therefore i∈I Ui = ( j ∈I Uj )K . 2 A quasi-uniformity U on X will be called nonsymmetric if it is not a uniformity on X, that is, if U = U −1 . Corollary 1. Suppose that U is a nonsymmetric quasi-uniformity such that U = UK . Then U ∧ U −1 = (U ∧ U −1 )K . Proof. This is a consequence of Remarks 1(b) and 1(c).
2
In the following we give some examples of classes of quasi-uniformities U on X for which U = UK . Our method of proof will be based on an idea explained in the next proposition. A general characterization of those quasi-uniformities U on a set X such that U = UK will be given in the next section (see Lemma 3). Let U1 and U2 be two quasi-uniformities on a set X. Then U1 and U2 will be called ∨-complementary provided that U1 ∨ U2 = D. Proposition 2. Let U be a quasi-uniformity on X. Then each quasi-uniformity V on X that is ∨-complementary to UK is ∨-complementary to U . (In particular, each complement of UK is a complement of U in the lattice (q(X), ⊆).) Proof. Suppose otherwise. Then there is a quasi-uniformity V on X such that V ∨ UK = D, but V ∨ U = D. By Zorn’s Lemma there is an anti-atom K of (q(X), ⊆) such that V ∨ U ⊆ K. Thus UK ⊆ K. But then D = V ∨ UK ⊆ K—a contradiction. We conclude that the first statement holds. The second statement immediately follows, since U ⊆ UK . 2 Definition 2. Let T be a preorder on X and let U be a quasi-uniformity on X. We say that T is U -slim provided that there is U ∈ U such that U ∩ T = Δ. Corollary 2. Let U be a quasi-uniformity on X. Then each preorder T on X that is UK -slim is U -slim. Proof. Otherwise the quasi-uniformity V = fil{T } would be ∨-complementary to UK , but not to U , contradicting Proposition 2. 2 Recall that a subset A of a quasi-uniform space (X, U) is called discrete if there is an entourage U ∈ U such that U ∩ (A × A) = ΔA (equivalently, if the subspace quasi-uniformity U|A is discrete). Corollary 3. Let U be a quasi-uniformity on X. Then UK and U have the same discrete subsets in X. Proof. Suppose that there are A ⊆ X and V ∈ UK such that V ∩ (A × A) = ΔA , but that U ∩ (A × A) = ΔA whenever U ∈ U . Set T = Δ ∪ (A × A). Then T is a preorder on X that is UK -slim but not U -slim. We have reached a contradiction to Corollary 2 and conclude that the statement holds. 2
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Corollary 4. Let U be a quasi-uniformity on X. Then for any disjoint A, B ⊆ X and any V ∈ UK there is U ∈ U such that U ∩ (A × B) ⊆ V . Proof. Suppose not. Set T = Δ ∪ [(A × B) \ V ]. Since A and B are disjoint, T is a preorder on X. It is UK -slim but not U -slim by our assumption, contradicting Corollary 2. 2 Corollary 5. Let U be a quasi-uniformity on X. Then UK belongs to the quasi-proximity class of U . (In particular, U and UK induce the same topology.) Proof. Otherwise there are A, B ⊆ X and V ∈ UK such that V ∩ (A × B) = ∅, but U ∩ (A × B) = ∅ whenever U ∈ U . In particular, A and B are disjoint. We have reached a contradiction to Corollary 4 and conclude that UK belongs to the quasi-proximity class of U . 2 Proposition 3. Each totally bounded quasi-uniformity U on a set X is the infimum of anti-atoms of (q(X), ⊆). Proof. In order to reach a contradiction we suppose that U ⊂ UK . We conclude by Corollary 5 that UK cannot be totally bounded, since only the coarsest quasi-uniformity of a quasi-proximity class is totally bounded [4, Theo−1 rem 1.33]. Hence either UK or UK is not hereditarily precompact [7, Lemma 1.1]. Consequently there are V ∈ UK and a sequence (xn )n∈ω in X such that xk ∈ / V (xn ) whenever n, k ∈ ω and n < k (or n > k). Define T1 = Δ ∪ {(xn , xk ): n k; n, k ∈ ω} (or T2 = Δ ∪ {(xn , xk ): n k; n, k ∈ ω}). Then T1 (or T2 ) is a UK -slim preorder. Note that T1 and T2 are not U -slim: Consider U ∈ U . There is a finite cover {Af : f ∈ F } of X such that Af × Af ⊆ U whenever f ∈ F . Hence there are e ∈ F and i, j ∈ ω such that i < j and xi , xj ∈ Ae . It follows that (xi , xj ) ∈ U ∩ T1 (and (xj , xi ) ∈ U ∩ T2 ). We have reached a contradiction to Corollary 2 and conclude that U = UK . 2 Indeed let us note that the preceding argument essentially shows the following: Corollary 6. If U is a quasi-uniformity on X that is hereditarily precompact, then UK is hereditarily precompact, too. Proof. Otherwise the proof above shows how to construct a preorder T on X that is UK -slim but not U -slim, which contradicts Corollary 2. 2 A quasi-uniformity U on X is called proximally fine if it is the finest quasi-uniformity inducing the quasiproximity δU . (But recall that in general a given quasi-proximity class need not possess a finest member [4, p. 25].) Corollary 7. Each proximally fine quasi-uniformity U on X is the infimum of anti-atoms of (q(X), ⊆). In particular, the fine quasi-uniformity of any topological space X is the infimum of anti-atoms of (q(X), ⊆). Proof. If U ⊂ UK and U is proximally fine, then U and UK must induce distinct quasi-proximities, which contradicts Corollary 5. 2 Corollary 8. Let U be a uniformity on X. Then the Hausdorff uniformities determined by UK and U induce the same topology on the set P0 (X) of nonempty subsets of X. Proof. Note that UK is a uniformity, because U is a uniformity. Since U ⊆ UK and since U and UK have the same discrete subsets and belong to the same quasi-proximity class by Corollaries 3 and 5, our assertion follows from a statement of Ward [13, Theorem 1]. 2 Lemma 1. Let U be a quasi-uniformity on X and suppose that V ∈ UK \ U . Let A ⊆ X be of minimal cardinality such that (A × A) ∩ (U \ V ) = ∅ whenever U ∈ U . If B ⊆ A and [(A × B) ∪ (B × A)] ∩ (U \ V ) = ∅ whenever U ∈ U , then we have |B| = |A|. (Note that A cannot be finite.)
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Proof. Assume first that A is finite. Then there must exist a point (a, b) ∈ A × A such that (a, b) ∈ ( U) \ V , since U is a filter. Consequently T = Δ ∪ {(a, b)} is a preorder on X that is UK -slim but not U -slim—a contradiction to Corollary 2. So A is necessarily infinite. In order to reach a contradiction, suppose now that there is a B ⊆ A such that |B| < |A| and [(A × B) ∪ (B × A)] ∩ (U \ V ) = ∅ whenever U ∈ U . By our assumption on A, there is a U0 ∈ U such that (B × B) ∩ (U0 \ V ) = ∅. Since (A × B) ∪ (B × A) = (B × B) ∪ (B × (A \ B)) ∪ ((A \ B) × B), we have that either (B × (A \ B)) ∩ (U \ V ) = ∅ or ((A \ B) × B) ∩ (U \ V ) = ∅ whenever U ∈ U . Let us consider the first case. (The second case can be treated similarly.) Set T = Δ ∪ [(B × (A \ B)) \ V ]. Then T is a preorder on X that is UK -slim but not U -slim—a contradiction to Corollary 2. We deduce that the statement holds. 2 Remark 2. Let us note that in Lemma 1 the infinite cardinal |A| is necessarily at most equal to the minimal cardinality of a base of the subspace quasi-uniformity U|A on A. Proposition 4. Each quasi-uniformity U on X with a linearly ordered base is the infimum of anti-atoms of (q(X), ⊆). In particular, each quasi-uniformity on X with a countable base is the infimum of anti-atoms of (q(X), ⊆). Proof. Let {Uβ : β < δ} be a decreasing base of U (that is, the sequence (Uβ )β<δ is well-ordered by inverse settheoretic inclusion). We can assume that the cardinal δ is equal to one, countably infinite or an uncountable regular cardinal (compare [10]). Suppose that there is V ∈ UK \ U . Let Y ⊆ X be of minimal cardinality such that (Y × Y ) ∩ (Uβ \ V ) = ∅ whenever β ∈ δ. As noted in the preceding lemma, Y is necessarily infinite. It also follows from Remark 2 that certainly |Y | δ. Indeed under our linearity condition, |Y | δ too: Suppose otherwise, that is, suppose |Y | < δ. Then δ is uncountable. Because Y is infinite, for each (y1 , y2 ) ∈ (Y × Y ) \ V there is a minimal β < δ, say β(y1 ,y2 ) , such that / Uβ(y1 ,y2 ) . Since sup{βz : z ∈ (Y × Y ) \ V } < δ by regularity of δ, we have reached the contradiction that (y1 , y2 ) ∈ V ∩ (Y × Y ) ∈ U|Y . Hence |Y | = δ. We now define two sequences (xβ )β∈δ and (yβ )β∈δ of points in Y inductively as follows: Assume that for some γ < δ and all β < γ we have already chosen xβ and yβ so that Aγ = {xβ : β < γ } and Bγ = {yβ : β < γ } are disjoint. Note that this condition of disjointness holds at limit ordinals provided that it holds at all the preceding ordinals. / Bγ , yγ ∈ / Aγ and Suppose that we can find (xγ , yγ ) ∈ [(Uγ \ V ) ∩ (Y × Y )] \ [(Bγ × Y ) ∪ (Y × Aγ )]. Then xγ ∈ xγ = yγ . Set Aγ +1 = {xβ : β < γ + 1} and Bγ +1 = {yβ : β < γ + 1}. Note that Aγ +1 and Bγ +1 are disjoint. The induction would stop if for some γ ∈ δ we had (Uγ \ V ) ∩ (Y × Y ) ⊆ [(Bγ × Y ) ∪ (Y × Aγ )]. But this clearly implies that (U \ V ) ∩ [[(Aγ ∪ Bγ ) × Y ] ∪ [Y × (Aγ ∪ Bγ )]] = ∅ whenever U ∈ U . Since Aγ ∪ Bγ has cardinality smaller than δ, the latter property cannot hold by definition of Y and Lemma 1. We conclude that this case cannot occur so that we can always continue the induction for all γ < δ. Finally we set T = Δ ∪ {(xγ , yγ ): γ < δ}. By the disjointness of Aδ and Bδ , T is a preorder on X. It is UK -slim but not U -slim—a contradiction to Corollary 2. Hence we have shown that U = UK . 2 A topological space X is called resolvable [6] if it has two disjoint dense subsets. Proposition 5. Each quasi-uniformity U on X such that τ (U s ) is resolvable is the infimum of anti-atoms of (q(X), ⊆). Proof. Let D1 , D2 be disjoint dense sets of (X, τ (U s )). In order to reach a contradiction suppose that there is a V ∈ UK such that U \ V 3 = ∅ whenever U ∈ U . Set T = Δ ∪ [(D1 × D2 ) \ V ]. Note that T is a preorder on X, since D1 and D2 are disjoint. Let us show that T is not U -slim: We shall prove that U 3 ∩ T = Δ whenever U ∈ U . Fix U ∈ U . There is (x, y) ∈ U \ V 3 . Then [V (x) × V −1 (y)] ∩ V = ∅. Since according to Corollary 5 the quasi-uniformities UK and U belong to the same quasi-proximity class and thus both induce the bitopological space (X, τ (U), τ (U −1 )), we can choose (d1 , d2 ) ∈ [[U −1 (x) ∩ V (x)] × [V −1 (y) ∩ U (y)]] ∩ [D1 × D2 ]. Then we have (d1 , x) ∈ U , (y, d2 ) ∈ U and (x, y) ∈ U . Therefore (d1 , d2 ) ∈ U 3 \ V . Hence (d1 , d2 ) ∈ U 3 ∩ T . We conclude that T is not U -slim. On the other hand, note that T is UK -slim, since clearly V ∩ T = Δ. We have reached a contradiction to Corollary 2 and conclude that U = UK . 2
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Proposition 6. Let U be a quasi-uniformity on X and let d be a quasi-pseudometric on X such that Ud ⊆ U , but Ud−1 U . Then there is an anti-atom K of (q(X), ⊆) such that U ⊆ K, but Ud−1 K and thus Ud−1 UK . Proof. For each n ∈ ω ∪ {−1} we set Qn = {(x, y) ∈ X × X: d(x, y) 2−n }. Without loss of generality we can assume that Q−1 / U. −1 ∈ For each n ∈ ω set Fn = Δ ∪ (Qn \ Q−1 −1 ). Define n = Fn0 ◦ · · · ◦ Fnk : k ∈ ω; n0 , . . . , nk ∈ ω; 2−n0 + · · · + 2−nk 2−n . F 2 n and F n whenever n ∈ ω. Consider V = fil{F n : n ∈ ω} ∨ U . Note that the quasi Observe that Fn ⊆ F n+1 ⊆ F n and uniformity V on X is distinct from the discrete quasi-uniformity because for each Q ∈ U , Q ∩ Fn ⊆ Q ∩ F −1 −1 Q ∩ Fn = Δ, since Q−1 ∈ / U and thus (Q ∩ Qn ) \ Q−1 = ∅. Therefore by Zorn’s Lemma there exists an anti-atom K of (q(X), ⊆) such that V ⊆ K. In particular, U ⊆ K. In order to see that Ud−1 K, we shall show that Q−1 / K. To this end, it will be sufficient to verify that F 0 ∩ Q−1 0 ∈ 0 = Δ, since Δ ∈ / K and F 0 ∈ K. In order to reach a contradiction assume that F 0 ∩ Q−1 0 = Δ: Then there are s ∈ ω \ {0} and xi ∈ X (i = 0, . . . , s) such that
(1) x0 = xs ;
s−1 −n j 1; and (2) for each j = 0, . . . , s − 1, (xj , xj +1 ) ∈ Qnj \ Q−1 j =0 2 −1 where nj ∈ ω and (3) (x0 , xs ) ∈ Q−1 0 .
Without loss of generality we could assume that xj = xj +1 . −1 By the triangle inequality it follows that (x0 , xs−1 ) ∈ Q0 . Hence (xs−1 , x0 ) ∈ Q−1 0 . Together with (x0 , xs ) ∈ Q0 −1 we conclude that (xs−1 , xs ) ∈ Q−1 −1 . Since also (xs−1 , xs ) ∈ Qns−1 \ Q−1 we have reached a contradiction. Hence −1 indeed F 0 ∩ Q0 = Δ. We have shown that there is an anti-atom K of (q(X), ⊆) such that U ⊆ K and Ud−1 ⊆ K. Therefore Ud−1 ⊆ UK . 2 Corollary 9. Let U be a quasi-uniformity on X. Then U −1 ∧ U = U −1 ∧ UK . Proof. Obviously U −1 ∧ U ⊆ U −1 ∧ UK . In order to prove the converse, recall that U −1 ∧ UK , as any quasiuniformity (see [4, Lemma 1.5]), can be written as the supremum of a family, say (Ud )d∈D , of quasi-pseudometric quasi-uniformities. Let d ∈ D. Then Ud ⊆ U −1 ∧ UK . If Ud ⊆ U , then applying Proposition 6 to U −1 we see that Ud−1 (U −1 )K and thus Ud UK —a contradiction. Hence Ud ⊆ U and Ud ⊆ U −1 ∧ U . Thus U −1 ∧ UK ⊆ U −1 ∧ U and the assertion is verified. 2 Corollary 10. Let U be a uniformity on X. Then any quasi-uniformity V on X such that U ⊆ V ⊆ UK is a uniformity. Proof. Suppose that there is a quasi-uniformity V on X such that V = V −1 and U ⊆ V ⊆ UK . Then U ⊆ V −1 ⊆ UK , because UK is a uniformity. But by Proposition 6 applied to V there is an anti-atom K on X such that V ⊆ K, but V −1 K. Then U ⊆ K and therefore UK ⊆ K. Consequently V −1 ⊆ UK ⊆ K—a contradiction. We conclude that the statement holds. 2 Corollary 11. Let U be any nonsymmetric quasi-uniformity on X and let V be a quasi-uniformity on X such that U ⊆ V ⊆ UK . Then V is nonsymmetric. Proof. Suppose otherwise, that is, that V = V −1 . Let K be any anti-atom of (q(X), ⊆) such that U ⊆ K. Then UK ⊆ K and hence V ⊆ K. Consequently V ⊆ K−1 and thus U ⊆ K−1 and U −1 ⊆ K. Applying Proposition 6 to U , we note that we have reached a contradiction and conclude that V is nonsymmetric. 2
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3. Some further results Remark 3. In [9, p. 7] it was pointed out that any separated uniformity U on an infinite set X with at least one nonisolated point x0 is not the infimum of anti-atoms in the lattice of uniformities on X: Clearly T = Δ ∪ [(X \ {x0 }) × (X \ {x0 })] is contained in any (uniform) anti-atom K finer than U although it certainly does not belong to U . (Indeed suppose that K is an anti-atom finer than U in the lattice of uniformities on X such that T ∈ / K. Then by maximality of K, there is a symmetric entourage K ∈ K such that K 2 ∩ [(X \ {x0 }) × (X \ {x0 })] = Δ. Thus K(x0 ) \ {x0 } can / U . But then contain at most one point, say a. Since U is separated, there is a symmetric U ∈ U such that (x0 , a) ∈ Δ = K ∩ U ∈ K—a contradiction.) On the other hand, Proposition 4 for instance shows that U = UK in the case that such a U has a linearly ordered base. Remark 4. Remark 3 above shows that an analogue to Corollary 5 does not hold in the lattice of uniformities on X. A potential way to construct an example of a quasi-uniformity U on X such that U ⊂ UK is sketched in our next observation. Example 1. Let U be a maximal element among the quasi-uniformities on X that do not contain some fixed preorder T . Then U is either an anti-atom of (q(X), ⊆) or is strictly coarser than UK . Proof. Note that U = D. If U is not an anti-atom of (q(X), ⊆), then any anti-atom K of (q(X), ⊆) finer than U contains T . Therefore T belongs to UK . Hence U and UK are distinct. 2 Problem 2. Is U as defined in Example 1 always an anti-atom of (q(X), ⊆)? Our next results will show a connection between our problem and concepts of finite dimensionality. Remark 5. Let U be a quasi-uniformity on X. Suppose that V ∈ UK \ U . Then (X × X) \ V cannot be written as the union of finitely many transitive relations. In particular, V cannot be written as the intersection of finitely many linear orders. Proof. We suppose that (X × X) \ V = i∈n Si where n ∈ ω and each Si is transitive. Set Ti = Δ ∪ Si for i ∈ n. Since U \ V = ∅ whenever U ∈ U , there is j ∈ n such that Tj ∩ (U \ V ) = ∅ whenever U ∈ U . Thus Tj is not U -slim, although it is UK -slim—a contradiction to Corollary 2. For the second statement, suppose that V = i∈n Li where each Li is a linear order on X. Then V = (X × X) \
i∈n Li where each Li = (X × X) \ Li is transitive, which is impossible by the result just proved. 2 In the preceding remark, V cannot be of the form {(x, y) ∈ X ×X: |f (x)−f (y)| < ε} for some function f : X → R and ε > 0, since clearly V = (X × X) \ (Z+ ∪ Z− ) where Z+ = {(x, y) ∈ X × X: f (x) f (y) + ε} and Z− = {(x, y) ∈ X × X: f (x) f (y) − ε}, and these two latter relations are transitive. Lemma 2. Let C = {Ca : a ∈ I } be a collection of pairwise disjoint subsets of a set X (with linearly ordered index set (I, )). Set A = C. Consider T0 = Δ ∪ {Ca × Ca : a ∈ I }. Then T0 is the complement of the union of six transitive relations on X. Proof. We assume that X is linearly ordered by . Set S0 = (X \ A) × A, S1 = A × (X \ A), S2 = {(x, y) ∈ (X \ A) × S = {Ca × (X \ A): x ≺ y}, S3 = {(x, y) ∈ (X \ A) × (X \ A): y ≺ x}, S4 = {Ca × Ca : a < a ; a, a ∈ I } and 5 Ca : a < a ; a, a ∈ I }. Then each Si where i ∈ 6 is a transitive relation on X and T0 = (X × X) \ ( i∈6 Si ). 2 Proposition 7. Let U be a quasi-uniformity on X, and let d be a quasi-pseudometric on X such that Ud ⊆ UK , but / U . (Here we set Un = {(x, y) ∈ X × X: d(x, y) < 2−n } and Uns = Un ∩ Un−1 whenever n ∈ ω.) U0 ∈ Suppose that each uniform cover of the uniform space (X, Uds ) has a refinement that consists of finitely many (uniformly) Uds -discrete collections. Then Ud−1 UK .
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Proof. Assume that X is linearly ordered. Consider the uniform cover {U2s (x): x ∈ X} of the uniform space (X, Uds ). By our assumption it has a refinement that can be written as the union of finitely many Uds -discrete collections Ci , where i ∈ n. We assume that these collections are linearly ordered. For each i ∈ n there is a ji ∈ ω \ 3 such that Ujsi ∈ Uds and Ujsi witnesses the uniform discreteness of Ci . Hence for each i ∈ n, {Ujsi (C): C ∈ Ci } is a collection of pairwise disjoint subsets of X. Set Ei = Δ ∪ {Ujsi (C) × Ujsi (C): C ∈ Ci } whenever i ∈ n. Then for each i ∈ n, the relation Ei is the complement in X × X of a union of six transitive relations on X by Lemma 2. Set U = i∈n Ujsi . We have U ⊆ i∈n Ei , since i∈n Ci is a cover of X. Furthermore, i∈n Ei ⊆ U0s ⊆ U0 . Hence i∈n Ei does not belong to U . Since the intersection of transitive relations is transitive, we conclude by the aforementioned structure of the complement of each Ei that i∈n Ei is the complement of a finite union of transitive relations on X. Thus it cannot belong to UK \ U according to Remark 5. It follows that i∈n Ei and thus U does not belong to UK , which implies the statement, because Ud ⊆ UK . 2 Corollary 12. Let U be a uniformity on a set X and let T be a preorder on X such that T ∈ / U . Then T ∈ / UK . Proof. Suppose T ∈ UK . Recall that since U is a uniformity, UK is a uniformity. Hence by our assumption fil{T ∩ T −1 } ⊆ UK . For the quasi-pseudometric d on X defined by d(x, y) = 0 if (x, y) ∈ T , and d(x, y) = 1 otherwise, we have U0 = T and Ud = fil{T }. However, then Ud−1 ⊆ UK which clearly contradicts Proposition 7, since the cover {(T ∩ T −1 )(x): x ∈ X} is a partition of X. Thus T ∈ / UK . 2 We next formulate a general characterization of those quasi-uniformities U on X that satisfy the equation U = UK . Lemma 3. A quasi-uniformity U on a set X is the infimum of anti-atoms in the lattice of quasi-uniformities on X if and only if for each quasi-pseudometric d on X such that Ud ⊆ U there exists a quasi-pseudometric d on X such that Ud ∨ U = D and Ud ∨ Ud = D. Proof. Suppose that the given condition holds. In order to reach a contradiction, assume that U ⊂ UK . Then there is a quasi-pseudometric d on X such that Ud ⊆ U , but Ud ⊆ UK . Therefore, by the stated condition there is a quasipseudometric d on X such that Ud ∨ U = D and Ud ∨ Ud = D. We can find an anti-atom K of (q(X), ⊆) such that Ud ∨ U ⊆ K. Then we have UK ⊆ K. Since D = Ud ∨ Ud ⊆ K, we have reached a contradiction. We conclude that the condition implies that U = UK . In order to establish the converse, assume that the stated condition does not hold. Then there is a quasi-pseudometric d on X such that Ud ⊆ U but that for any quasi-pseudometric d on X such that Ud ∨ U = D, we have that Ud ∨ Ud = D. Consider an arbitrary anti-atom K of (q(X), ⊆) such that U ⊆ K. In order to reach a contradiction, we assume that K ∨ Ud = D. This means that there is a quasi-pseudometric d on X such that Ud ⊆ K, Ud ∨ Ud = D and Ud ∨ U = D—a contradiction. Therefore K ∨ Ud = D and thus Ud ⊆ K, since K is an anti-atom. We have shown that Ud ⊆ UK . Thus U ⊂ UK . We conclude that U = UK implies the given condition. Hence the statement of the lemma holds. 2 Remark 6. Note that the lemma above also yields a characterization of those uniformities on a set X that are infima of families of anti-atoms in the lattice of uniformities on X: Just replace quasi-pseudometrics everywhere by pseudometrics in its formulation. Remark 7. It seems natural to distinguish between the proximally discrete and the proximally nondiscrete anti-atoms when studying Problem 1 (see [3]). For instance, it is a trivial observation that for any quasi-uniformity U on X the infimum of all proximally discrete anti-atoms of (q(X), ⊆) above it is finer than Dω . Remark 8. In the following, for any filters F and G on X, we define the quasi-uniformity UF ,G = fil{Δ∪(F ×G): F ∈ F, G ∈ G} on X. It follows from [3, Section 3] that for each anti-atom K of (q(X), ⊆) there are uniquely defined ultrafilters F and G on X such that F ∩ G = ∅ and UF ,G ⊆ K. (Note that in [3] these ultrafilters F and G were denoted by pr1 K and pr2 K .)
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Fix now some ultrafilters F and G on X such that F ∩ G = ∅. If U is a quasi-uniformity on X, then there are anti-atoms finer than UF ,G above it if and only if U ∨ UF ,G = D. In particular, each quasi-uniformity U = D that is finer than UF ,G clearly has only anti-atoms K of (q(X), ⊆) above it that satisfy pr1 K = F and pr2 K = G. Example 2. Let x ∈ X and F = fil{{x}}, and let G be a free ultrafilter on X. Then UF ,G ∧ UG ,F ⊂ UG ,G . Is known [3, Theorem 1] that UF ,G and UG ,F are anti-atoms of (q(X), ⊆), but that UG ,G is not an anti-atom of (q(X), ⊆) [3, Corollary 5]. Note that if U is an anti-atom of the lattice of uniformities on X that contains the uniformity UG ,G , then any anti-atom K of (q(X), ⊆) containing U satisfies UG ,G ⊆ K ∧ K−1 , where the latter infimum is equal to U . Proposition 8. Each quasi-uniformity U on X that is finer than UF ,G , where F and G are distinct ultrafilters on X, is the infimum of anti-atoms of (q(X), ⊆). Proof. There exists A ⊆ X such that A ∈ F and X \ A ∈ G. Let V ∈ UK . By Corollary 4 there is U ∈ U such that U ∩ [Δ ∪ (A × (X \ A))] ⊆ V . But Δ ∪ (A × (X \ A)) ∈ U . Thus V ∈ U and we have shown that U = UK . 2 Problem 3. Does Proposition 8 also hold if F and G are the same free ultrafilter on X? (It follows from Proposition 10 below and Remark 1(c) that an affirmative answer to this question would yield a positive solution to our Problem 1.) Proposition 9. Let U be a quasi-uniformity on a set X. Then U ∨ Dω = set of ultrafilters on X.
F (U
∨ UF ,F ), where F runs through the
Proof. The inclusion ⊆ is obvious. Let V ∈ F (U ∨ UF ,F ). Then for each ultrafilter F on X there are UF ∈ U and AF ∈ F such that UF ∩ (Δ ∪ (AF × AF )) ⊆ V . If {X \ AF : F is an ultrafilter on X} has the finite intersection property, this collection is contained in some ultrafilter G on X; however, this means that ∅ = (X \ AG ) ∩ AG ∈ G and we have reached a contradiction. Thus there is a finite collection E of ultrafilters F on X such that F ∈E AF = X. Consequently ( F ∈E UF ) ∩ F ∈E (AF × AF ) ⊆ V . Therefore V ∈ U ∨ Dω . We conclude that the stated equality holds. 2 Proposition 10. Let U be a quasi-uniformity on a set X. Then U = the set of ultrafilters on X.
(F ,G ) (U
∨ UF ,G ), where F and G run through
Proof. The inclusion ⊆ is evident. Let V ∈ (F ,G ) (U ∨ UF ,G ). According to Proposition 9 there are U0 ∈ U and a finite cover {Ci : i ∈ n} (where n ∈ ω) of X such that U0 ∩ ( i∈n (Ci × Ci )) ⊆ V . Clearly, without loss of generality we can assume that {Ci : i ∈ n} is a partition of X. Furthermore, for each pair (F, G) of distinct ultrafilters F and G on X there are U(F ,G ) ∈ U and disjoint sets F(F ,G ) ∈ F and G(F ,G ) ∈ G such that [Δ ∪ (F(F ,G ) × G(F ,G ) )] ∩ U(F ,G ) ⊆ V . Fix F . Determine i ∈ n such that Ci ∈ F and denote it by CF . Clearly HF := fil{X \ G(F ,G ) : G is an ultrafilter distinct from F on X} ⊆ F , because F(F ,G ) ⊆ X \ G(F ,G ) whenever G is an ultrafilter distinct from F on X. If HF ⊂ F , then HF ⊆ G0 for some ultrafilter G0 on X distinct from F , which leads to the contradiction that Therefore ( G ∈EF X \ G(F ,G ) ) ⊆ CF for some finite subset EF of (X \ G(F ,G0 ) ) ∩ G(F ,G0 ) ∈ G0 . Thus HF = F . ultrafilters G distinct from F on X. Set LF = ( G ∈EF F(F ,G ) ) ∩ CF . Consider now {X \ LF : F is an ultrafilter on X}. If the latter collection has the finite intersection property, then it is contained in an ultrafilter F on X. But then (X \ LF ) ∩ LF ∈ F —a contradiction. Thus there finite collection I is a of ultrafilters F on X such that F ∈I LF = X. Note now that X × X = [ i∈n (Ci × Ci )] ∪ [ F ∈I ( G ∈EF (F(F ,G ) × G(F ,G ) ))]: Indeed let (x, y) ∈ X × X. Then x ∈ Ci and y ∈ Cj for some i, j ∈ n. If i = j , then (x, y) ∈ Ci × Ci . So let us consider the case where i = j . Of course, x ∈ LF for some F ∈ I and so x ∈ Ci ∩ CF and thus Ci = CF . Furthermore y∈ / Ci = CF . Consequently y ∈ G(F ,G ) for some G ∈ EF . We conclude that (x, y) ∈ F(F ,G ) × G(F ,G ) where F ∈ I and G ∈ EF . Hence our auxiliary set-theoretic equality is proved. From this equality and our definition of the entourages U0 and U(F ,G ) it now follows that U0 ∩ F ∈I ( G ∈EF U(F ,G ) ) ⊆ V . Therefore V ∈ U . We conclude that the statement holds. 2
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Corollary 13. Let U be a quasi-uniformity on X. Then U = F (U ∨ UF ,F ) ∧ {K: K is a proximally nondiscrete anti-atom of (q(X), ⊆) finer than U}, where F runs through the set of ultrafilters on X. Proof. This assertion follows from the fact that any anti-atom finer than UF ,G with distinct ultrafilters F and G on X is proximally nondiscrete (see [3, Section 3]) and Propositions 8 and 10 proved above. 2 In the light of the difficulty of Problem 1 it seems appropriate to look for a not too large set QX of quasi-uniformities on X that contains each anti-atom of (q(X), ⊆) as a member and where we can prove that each quasi-uniformity on X can be written as the infimum of a subset of QX . Our next result shows that such a natural set QX indeed exists. In order to define the set QX we need the following definitions: Let H be any ultrafilter on X × X. Define the filter as a subset. Of course, = {Δ ∪ H : H ∈ H} on X × X. Let (H) q be the finest quasi-uniformity on X contained in H H q = D if and only if Δ ∈ H. (H) / H. Finally set QX equal to the set of Now set HX equal to the set of all the ultrafilters H on X × X such that Δ ∈ q where H belongs to HX . all quasi-uniformities of the form (H) Remark 9. Using the notation just introduced, we remark that it is known [3, Theorem 1] that if H is an ultrafilter on is a proximally nondiscrete anti-atom of (q(X), ⊆). If that condition regarding X × X such that pr1 H = pr2 H, then H will necessarily be a quasi-uniformity. the projections is not satisfied, there seems to be no reason to believe that H q. But note that for any ultrafilter H on X × X such that Δ ∈ / H, we have Upr1 H,pr2 H ⊆ (H) Proposition 11. (a) Each quasi-uniformity U on a set X can be represented as the intersection of quasi-uniformities belonging to QX . q , where H is any ultrafilter finer than the filter K = (b) Each anti-atom K of (q(X), ⊆) is of the form (H) fil{K \ Δ: K ∈ K} on X × X. Hence K ∈ QX . Proof. (a) Since Consider the filter U = fil{U \ Δ: U ∈ U} on X × X. ∧∅ = D, we can assume that U is nondiscrete.
H ∈ HX such that U ⊆ H} an ultrafilter on X × X such that U ⊆ H}. Therefore U = {H: Thus U = {H: H is
and consequently U = {(H)q : H ∈ HX such that U ⊆ H}. q for any ultrafilter H on X × X such that K ⊆ H by (b) If K is an anti-atom of (q(X), ⊆), then clearly K = (H) the maximality of K. 2 q )K hold for each ultrafilter H on X × X such that Δ ∈ q = ((H) / H? Problem 4. Does (H) (Because of Proposition 11(a) and Remark 1(c) an affirmative answer to this question would yield a positive solution to Problem 1.) Finally we note that the studied property of quasi-uniformities is preserved under finite unions, and we then apply this result to our Proposition 5 dealing with resolvability that we have presented in the preceding section. Proposition 12. Let U be a quasi-uniformity on a set X. Suppose that A ⊆ X such that (U|A)K = U|A and (U|(X \ A))K = U|(X \ A). Then U = UK . Proof. For any B, C ⊆ X, we set UB,C = U ∨ fil{Δ ∪ (B × C)}. Note first that U = UA,A ∩ UA,X\A ∩ UX\A,A ∩ UX\A,X\A . This equality is established in a similar fashion as the equalities in the proofs of Propositions 9 and 10 with the help of the fact that X × X = (A × A) ∪ (A × (X \ A)) ∪ ((X \ A) × A) ∪ ((X \ A) × (X \ A)). Applying Corollary 4 to the quasi-uniformity UA,X\A and the disjoint subsets A and X \ A of X we immediately see that (UA,X\A )K = UA,X\A . Similarly we conclude that (UX\A,A )K = UX\A,A . Note next that UA,A |A = U|A. Consider any quasi-pseudometric d on X such that Ud UA,A . Then there is m ∈ ω such that [U ∩ (A × A)] \ Um = ∅ whenever U ∈ U . (Here, as before, Un = {(x, y) ∈ X × X: d(x, y) < 2−n } whenever n ∈ ω.) Therefore Ud |A ⊆ UA,A |A. Since (U|A)K = U|A, by Lemma 3 there is a quasi-pseudometric q on A that is bounded by 1 such that Uq ∨ (Ud |A) = D|A and Uq ∨ (U|A) = D|A. Set s(x, y) = q(x, y) if x, y ∈ A, s(x, y) = 0 if x = y and x ∈ X \ A, and s(x, y) = 1 otherwise. Then s is a quasi-pseudometric on X. Furthermore, Us ∨ Ud = D and
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Us ∨ UA,A = D. We conclude that (UA,A )K = UA,A by Lemma 3. Similarly one shows that (UX\A,X\A )K = UX\A,X\A . The statement then follows from Remark 1(c). 2 A topological space that is not resolvable is called irresolvable. It is known that each topological space is the union of two disjoint subspaces C and U where C is closed and resolvable and U is open and hereditarily irresolvable, that is, every nonempty subspace of X is irresolvable [1, p. 292]. Corollary 14. A quasi-uniformity U on X satisfies U = UK if it satisfies U|U = (U|U )K , where U is the hereditarily irresolvable part in the aforementioned representation of the topological space (X, τ (U s )). Proof. By Proposition 5 we have (U|C)K = U|C where C is the resolvable part of the space (X, τ (U s )) mentioned in the representation above. The result follows from Proposition 12 and our hypothesis regarding U|U . 2 Problem 5. For i ∈ {0, 1}, let Ui be two quasi-uniformities on a set X such that Ui = (Ui )K . Does U0 ∨ U1 = (U0 ∨ U1 )K hold? References [1] G. Bezhanishvili, R. Mines, P.J. Morandi, Scattered, Hausdorff-reducible, and hereditarily irresolvable spaces, Topology Appl. 132 (2003) 291–306. [2] J. Cao, I.L. Reilly, S. Romaguera, Some properties of quasi-uniform multifunction spaces, J. Austral. Math. Soc. (Ser. A) 64 (1998) 169–177. [3] E.P. de Jager, H.-P.A. Künzi, Atoms, anti-atoms and complements in the lattice of quasi-uniformities, Topology Appl. 153 (16) (2006) 3140– 3156. [4] P. Fletcher, W.F. Lindgren, Quasi-Uniform Spaces, Dekker, Basel, 1982. [5] L.M. García-Raffi, S. Romaguera, E.A. Sánchez-Pérez, The bicompletion of an asymmetric normed linear space, Acta Math. Hungar. 97 (2002) 183–191. [6] E. Hewitt, A problem of set-theoretic topology, Duke Math. J. 10 (1943) 309–333. [7] H.-P.A. Künzi, Functorial admissible quasi-uniformities on topological spaces, Topology Appl. 43 (1992) 27–36. [8] H.-P.A. Künzi, Nonsymmetric distances and their associated topologies: About the origins of basic ideas in the area of asymmetric topology, in: C.E. Aull, R. Lowen (Eds.), Handbook of the History of General Topology, vol. 3, Kluwer, Dordrecht, 2001, pp. 853–968. [9] J. Pelant, J. Reiterman, V. Rödl, P. Simon, Ultrafilters on ω and atoms in the lattice of uniformities, I and II, Topology Appl. 30 (1988) 1–17 and 107–125. [10] H.-C. Reichel, Basis properties of topologies compatible with (not necessarily symmetric) distance-functions, Gen. Topology Appl. 8 (1978) 283–289. [11] J. Rodríguez-López, S. Romaguera, The relationship between the Vietoris topology and the Hausdorff quasi-uniformity, Topology Appl. 124 (2002) 451–464. [12] S. Romaguera, M. Sanchis, M. Tkachenko, Free paratopological groups, in: Proceedings 17th Summer Conference on Topology and its Applications, Topology Proc. 27 (2003) 613–640. [13] A.J. Ward, On H -equivalence of uniformities (the Isbell–Smith problem), Pacific J. Math. 22 (1967) 189–196.